Problem: Is ${644029}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {644029}= &&{6}\cdot100000+ \\&&{4}\cdot10000+ \\&&{4}\cdot1000+ \\&&{0}\cdot100+ \\&&{2}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {644029}= &&{6}(99999+1)+ \\&&{4}(9999+1)+ \\&&{4}(999+1)+ \\&&{0}(99+1)+ \\&&{2}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {644029}= &&\gray{6\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {6}+{4}+{4}+{0}+{2}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${644029}$ is divisible by $3$ if ${ 6}+{4}+{4}+{0}+{2}+{9}$ is divisible by $3$ Add the digits of ${644029}$ $ {6}+{4}+{4}+{0}+{2}+{9} = {25} $ If ${25}$ is divisible by $3$ , then ${644029}$ must also be divisible by $3$ ${25}$ is not divisible by $3$, therefore ${644029}$ must not be divisible by $3$.